Chapter 17.3, Problem 17.111P

A uniform slender rod of length L is dropped onto rigid supports at A and B. Because support B is slightly lower than support A, the rod strikes A with a velocity ${\bar{v}}_1$ before it strikes B. Assuming perfectly elastic impact at both A and B, determine the angular velocity of the rod and the velocity of its mass center immediately after the rod (a) strikes support A, (b) strikes support B, (c) again strikes support A.

Fig. P17.111

To determine

Find the angular velocity of the rod and the velocity of its mass center immediately after the rod strikes support A.

Answer to Problem 17.111P

The angular velocity of the rod when mass center immediately after the rod strikes support A is $\underset{\_}{\frac{3v_1}{L}}$.

The velocity of the rod when mass center immediately after the rod strikes support A $\underset{\_}{\frac{1}{2}{v}_1}$.

Explanation of Solution

Given information:

The length of uniform slender rod is L.

The mass of uniform slender rod is m.

The initial velocity of bar before it strikes B is ${\overline{v}}_1$.

Calculation:

Write the equation of centroidal moment of inertia $\left(\overline{I}\right)$ of the uniform slender rod.

$\overline{I}=\frac{1}{12}m{L}^2$

The impact is perfectly elastic at both A and B. Therefore, the coefficient of restitution is one $\left(e=1\right)$.

Write the impact condition for the given system after the rod strikes support A.

${\left(v_A\right)}_2=v_1\uparrow$

Here, ${\left(v_A\right)}_2$ is the velocity of the uniform slender rod after the rod strikes support A.

Write the equation of velocity of rod $\left({\overline{v}}_2\right)$ at its mass center after the rod strikes support A.

${\overline{v}}_2=\frac{L}{2}{\omega }_2-{\left(v_A\right)}_2$

Here, ${\omega }_2$ is the angular velocity of the uniform slender rod after rod strikes support A.

Substitute $v_1$ for ${\left(v_A\right)}_2$.

${\overline{v}}_2=\frac{L}{2}{\omega }_2-v_1$ (1)

Consider the impulse and momentum principle.

Sketch the impulse and momentum diagram for the first impact at A of the bar as shown in Figure (1).

Here, ${\displaystyle \int Adt}$ is the magnitude of the impulse acting at support A and $dt$ is the time interval.

Refer Figure (1).

Take moment about A (positive sign in clockwise direction).

$\begin{array}{l}m{v}_1\frac{L}{2}+0=m{\overline{v}}_2\frac{L}{2}+\overline{I}{\omega }_2\\ m{v}_1\frac{L}{2}=m{\overline{v}}_2\frac{L}{2}+\overline{I}{\omega }_2\end{array}$

Substitute $\frac{1}{12}m{L}^2$ for $\overline{I}$ and $\frac{L}{2}{\omega }_2-v_1$ for ${\overline{v}}_2$

$\begin{array}{c}m{v}_1\frac{L}{2}=m\left(\frac{L}{2}{\omega }_2-v_1\right)\frac{L}{2}+\left(\frac{1}{12}m{L}^2\right){\omega }_2\\ =m\frac{L^2}{4}{\omega }_2-m{v}_1\frac{L}{2}+\frac{1}{12}m{L}^2{\omega }_2\end{array}$

Simplify the Equation:

$\begin{array}{l}m{v}_1\frac{L}{2}+m{v}_1\frac{L}{2}=\frac{1}{4}m{L}^2{\omega }_2+\frac{1}{12}m{L}^2{\omega }_2\\ \frac{2}{2}m{v}_{1}L=\frac{4}{12}m{L}^2{\omega }_2\\ m{v}_{1}L=\frac{1}{3}m{L}^2{\omega }_2\\ {\omega }_2=\frac{3m{v}_{1}L}{m{L}^2}\\ {\omega }_2=\frac{3v_1}{L}\end{array}$

Thus, the angular velocity of the rod and the velocity of its mass center immediately after the rod strikes support A is $\underset{\_}{\frac{3v_1}{L}}$.

Find the velocity of the rod when mass center immediately after the rod strikes support A using Equation (1).

${\overline{v}}_2=\frac{L}{2}{\omega }_2-v_1$

Substitute $\frac{3v_1}{L}$ for ${\omega }_2$.

$\begin{array}{c}{\overline{v}}_2=\frac{L}{2}\left(\frac{3v_1}{L}\right)-v_1\\ =\frac{3v_1}{2}-v_1\\ =\frac{1}{2}{v}_1\end{array}$

Thus, the velocity of the rod when mass center immediately after the rod strikes support A $\underset{\_}{\frac{1}{2}{v}_1}$.

To determine

Find the angular velocity of the rod and the velocity of its mass center immediately after the rod strikes support B.

Answer to Problem 17.111P

The angular velocity of the rod when mass center immediately after the rod strikes support B is $\underset{\_}{\frac{3v_1}{L}}$.

The velocity of the rod when mass center immediately after the rod strikes support B is $\underset{\_}{\frac{1}{2}{v}_1}$.

Explanation of Solution

Calculation:

Write the impact condition for the given system after the rod strikes support B.

${\left(v_B\right)}_3=2v_1\uparrow$

Here, ${\left(v_A\right)}_2$ is the velocity of the uniform slender rod after the rod strikes support B.

Write the equation of velocity of rod $\left({\overline{v}}_3\right)$ at its mass center after the rod strikes support B.

${\overline{v}}_3={\left(v_B\right)}_3-\frac{L}{2}{\omega }_3$

Here, ${\omega }_3$ is the angular velocity of the uniform slender rod after rod strikes support B.

Substitute $2v_1$ for ${\left(v_B\right)}_3$.

${\overline{v}}_3=2v_1-\frac{L}{2}{\omega }_3$ (2)

Consider the impulse and momentum principle.

Sketch the impulse and momentum diagram for impact at B of the bar as shown in Figure (2).

Here, ${\displaystyle \int Bdt}$ is the magnitude of the impulse acting at B.

Refer Figure (2).

Take moment about B (positive sign in clockwise direction).

$\begin{array}{l}-m{\overline{v}}_2\frac{L}{2}+\overline{I}{\omega }_2+0=m{\overline{v}}_3\frac{L}{2}-\overline{I}{\omega }_3\\ -m{\overline{v}}_2\frac{L}{2}+\overline{I}{\omega }_2=m{\overline{v}}_3\frac{L}{2}-\overline{I}{\omega }_3\end{array}$

Substitute $\frac{1}{12}m{L}^2$ for $\overline{I}$, $\frac{1}{2}{v}_1$ for ${\overline{v}}_2$, $\frac{3v_1}{L}$ for ${\omega }_2$, and $2v_1-\frac{L}{2}{\omega }_3$ for ${\overline{v}}_3$.

$\begin{array}{l}-m\left(\frac{1}{2}{v}_1\right)\frac{L}{2}+\left(\frac{1}{12}m{L}^2\right)\frac{3v_1}{L}=m\left(2v_1-\frac{L}{2}{\omega }_3\right)\frac{L}{2}-\left(\frac{1}{12}m{L}^2\right){\omega }_3\\ -\frac{1}{4}m{v}_{1}L+\frac{1}{4}m{v}_{1}L=2m{v}_1\frac{L}{2}-m\frac{L^2}{4}{\omega }_3-\frac{1}{12}m{L}^2{\omega }_3\\ 0=2m{v}_1\frac{L}{2}-m\frac{L^2}{4}{\omega }_3-\frac{1}{12}m{L}^2{\omega }_3\end{array}$

$\begin{array}{l}2m{v}_1\frac{L}{2}=m\frac{L^2}{4}{\omega }_3+\frac{1}{12}m{L}^2{\omega }_3\\ m{v}_{1}L=\frac{1}{3}m{L}^2{\omega }_3\\ {\omega }_3=\frac{3m{v}_{1}L}{m{L}^2}\\ {\omega }_3=\frac{3v_{1}L}{L}\end{array}$

Thus, the angular velocity of the rod when mass center immediately after the rod strikes support B is $\underset{\_}{\frac{3v_1}{L}}$.

Find the velocity of the rod when mass center immediately after the rod strikes support B using Equation (2).

${\overline{v}}_3=2v_1-\frac{L}{2}{\omega }_3$

Substitute $\frac{3v_1}{L}$ for ${\omega }_3$.

$\begin{array}{c}{\overline{v}}_3=2v_1-\frac{L}{2}\left(\frac{3v_1}{L}\right)\\ =2v_1-\frac{3v_1}{2}\\ =\frac{1}{2}{v}_1\end{array}$

Thus, the velocity of the rod when mass center immediately after the rod strikes support B $\underset{\_}{\frac{1}{2}{v}_1}$.

To determine

Find the angular velocity of the rod and the velocity of its mass center immediately after the rod strikes support A again.

Answer to Problem 17.111P

The angular velocity of the rod when mass center immediately after the rod strikes support A again is $\underset{\_}{0}$.

The velocity of the rod when mass center immediately after the rod strikes support A again is $\underset{\_}{v_1}$.

Explanation of Solution

Calculation:

Write the impact condition for the given system after the rod again strikes support A.

${\left(v_A\right)}_4=v_1\uparrow$

Here, ${\left(v_A\right)}_4$ is the velocity of the uniform slender rod the rod again strikes support A.

Write the equation of velocity of rod $\left({\overline{v}}_4\right)$ at its mass center after the rod again strikes support.

${\overline{v}}_4={\left(v_A\right)}_4+\frac{L}{2}{\omega }_4$

Here, ${\omega }_4$ is the angular velocity of the uniform slender rod after rod again strikes support A.

Substitute $v_1$ for ${\left(v_A\right)}_4$.

${\overline{v}}_4=v_1+\frac{L}{2}{\omega }_4$ (3)

Consider the impulse and momentum principle.

Sketch the impulse and momentum diagram for second impact at A of the bar as shown in Figure (3).

Refer Figure (3).

Take moment about A (positive sign in clockwise direction).

$\begin{array}{l}m{\overline{v}}_3\frac{L}{2}+\overline{I}{\omega }_3+0=m{\overline{v}}_4\frac{L}{2}+\overline{I}{\omega }_4\\ m{\overline{v}}_3\frac{L}{2}+\overline{I}{\omega }_3=m{\overline{v}}_4\frac{L}{2}+\overline{I}{\omega }_4\end{array}$

Substitute $\frac{1}{12}m{L}^2$ for $\overline{I}$, $\frac{1}{2}{v}_1$ for ${\overline{v}}_3$, $\frac{3v_1}{L}$ for ${\omega }_3$, and $v_1+\frac{L}{2}{\omega }_4$ for ${\overline{v}}_4$.

$\begin{array}{l}m\left(\frac{1}{2}{v}_1\right)\frac{L}{2}+\left(\frac{1}{12}m{L}^2\right)\left(\frac{3v_1}{L}\right)=m\left(v_1+\frac{L}{2}{\omega }_4\right)\frac{L}{2}+\left(\frac{1}{12}m{L}^2\right){\omega }_4\\ \frac{1}{4}m{v}_{1}L+\frac{1}{4}m{v}_{1}L=\frac{1}{2}m{v}_{1}L+m\frac{L^2}{4}{\omega }_4+\frac{1}{12}m{L}^2{\omega }_4\\ \frac{1}{4}m{v}_{1}L+\frac{1}{4}m{v}_{1}L-\frac{1}{2}m{v}_{1}L=m\frac{L^2}{4}{\omega }_4+\frac{1}{12}m{L}^2{\omega }_4\\ 0=\frac{1}{4}m{L}^2{\omega }_4\\ {\omega }_4=0\end{array}$

Thus, the angular velocity of the uniform slender rod after rod again strikes support A support B is $\underset{\_}{0}$.

Find the velocity of rod at its mass center after the rod again strikes support A using Equation (3).

${\overline{v}}_4=v_1+\frac{L}{2}{\omega }_4$

Substitute 0 for ${\omega }_4$.

$\begin{array}{c}{\overline{v}}_4=v_1+0\\ =v_1\end{array}$

Thus, the velocity of rod at its mass center after the rod again strikes support A is $\underset{\_}{v_1}$.